|
|
|
|
Suppose that it is noon. You jump up, and you fall back towards the earth. You do not fall into the sun. No surprise: although the sun is much more massive than the earth, it is much further away, so, near the surface of the earth, the gravitational field of the earth is much stronger than that of the sun.
So, how high would you have to go before the sun's gravitational attraction would win? At what point, on a line between the earth and the sun, are the two gravitational forces exerted by the sun and the earth equal and opposite? Let's say it is at a point d from the centre of the earth, and D from the centre of the sun. Obviously d is much less than D because the sun's mass (Ms = 1.99 x 1030 kg) is much greater than the mass of the earth (Me = 5.98 x 1024 kg). But just how far away is that point at d? For the calculation, we shall need the distance between the earth and the sun, which on average is re = 1.50 x 108 km.
Let's draw a diagram -- not to scale at first, because we still haven't done the calculation: let's put a mass m at distance d from the moon and D from the sun.
|
© 2005. J.Wolfe@unsw.edu.au, phone 61-
2-9385 4954 (UT + 10, +11 Oct-Mar). School of Physics, University of New South Wales, Sydney, Australia.
|
|
|